Question

con resolución
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Answer 1

Respuesta:

$f'(4) =-\dfrac{169 }{27 }$

Explicación paso a paso:

Primero derivamos la expresión :

$f'(x) = \dfrac{(-6x)(\sqrt{2x+1} )-(\dfrac{2}{2\sqrt{2x+1} } )(1-3x^{2} )}{(\sqrt{2x+1})^{2} }$

$f'(x) = \dfrac{-6x\sqrt{2x+1} -\dfrac{1-3x^{2} }{\sqrt{2x+1} } }{2x+1 }$

$f'(x) = \dfrac{-6x(2x+1)-(1-3x^{2}) }{(2x+1)(\sqrt{2x+1} ) }$

$f'(x) = \dfrac{-12x^{2} -6x-1+3x^{2} }{(2x+1)(\sqrt{2x+1} ) }=\dfrac{-9x^{2} -6x-1}{(2x+1)(\sqrt{2x+1} ) }=-\dfrac{(3x +1)^{2} }{(2x+1)(\sqrt{2x+1} ) }$

Calculamos la deriva para x = 4 :

$f'(4) =-\dfrac{(3(4) +1)^{2} }{(2(4)+1)(\sqrt{2(4)+1} ) }=-\dfrac{13^{2} }{9\sqrt{9} }=-\dfrac{169 }{(9)(3) }=-\dfrac{169 }{27 }$